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This PDF gives the following definition of cosets for lattices:

enter image description here

However, I tried to picture a lattice $L$ with basis $(2,0),(0,2)$ and $\mathbb{Z}^2$ in order to take the quotient $\mathbb{Z}^2/L$. A coset is a shift of $L$ by $v\in\mathbb{Z}^2$. If I take $v=(0,1), (0,2), (0,3), ...,(0,n)$, they all represent the same coset because I'm shifting to the same structure, right? There is also another coset, the one represented by $(1,0)$. There's also the coset represented by $(0,0)$ So there are just $3$ cosets?

I don't know if this is right, because $(1,1) + L$ is also a shift, but in which coset is $(1,1)$ a representant?

My book says that the number of cosets relates to the determinant of the matrix of basis for the lattice $L$ like this:

$$|\mathbb{Z}^n/L| = |det \ B|$$

How is this relation given? It's not immediate for me. Maybe I understood the concept wrong.

Eran
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Poperton
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    Are you sure the tag 'lattice-orders' is appropriated here? Maybe you mean 'integer-lattices'. Notice that it is always more likely to get good answers (in this case, perhaps you already have one) if you tag the questions appropriately. – amrsa Jun 03 '18 at 08:48

1 Answers1

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The four cosets of $\mathbb{Z}^2/L$ in your example are $$L,\quad (0,1)+L,\quad (1,0)+L,\quad (1,1)+L.$$ Explicitly, $$L = \{\lambda(2,0) + \mu(0,2)\;:\; \lambda,\mu\in\mathbb{Z}\} = \{(2\lambda, 2\mu)\;:\; \lambda,\mu\in\mathbb{Z}\},$$ $$(0,1) + L = \{(0,1) + \lambda(2,0) + \mu(0,2)\;:\; \lambda,\mu\in\mathbb{Z}\} = \{(2\lambda, 2\mu+1)\;:\; \lambda,\mu\in\mathbb{Z}\},$$ etc. You should try drawing these four cosets: for each one, draw a pair of coordinate axes and a dotted grid representing $\mathbb{Z}^2$, and then circle the dots that are included.

(However, there are a bunch of different ways of writing all of these elements. For instance, the set $$(5,8) + L = \{(5,8) + \lambda'(2,0) + \mu'(0,2)\;:\; \lambda',\mu'\in\mathbb{Z}\} = \{(2\lambda'+5, 2\mu'+8)\;:\; \lambda',\mu'\in\mathbb{Z}\}$$ can easily be seen to be equal to $(1,0)+L$ by reparametrising: set $\lambda' = \lambda-2, \mu' = \mu-4$. Alternatively, just draw it - you'll see that the pictures end up the same.)

Finally, the determinant of a matrix $B$ can be calculated as follows: take the unit cube, apply $B$ to it, and the volume of the resulting parallelepiped is $\det(B)$. (If you've never seen this before, try to prove it explicitly when $B = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Draw a picture of the unit square, and then a picture of the parallelogram you get when you apply $B$ to it, and calculate its area.)

Billy
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  • Thank you, just one thing: why the determinant is equal to the number of cosets? – Poperton Jun 03 '18 at 00:29
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    Take the unit square $X = [0,1)\times[0,1)$ (or unit cube in higher dimensions), and apply $B$ to it. The cosets of $L$ are precisely $(a,b) + L$, where $(a,b)$ is an integer point that belongs to $B(X)$. The number of those points is the area of $B(X)$, which is $\det(B)$. – Billy Jun 03 '18 at 10:38